Answer: C. 
would increase and 
would decrease.
Explanation:
initially conc. c 0 0
At eqm. 
The expression for dissociation constant is,
for weak acid, 
is very very small , the expression will be,
As 
is directly proportional to concentration, the value of would increase on increasing the concentration.
Thus if would increase, 
would decrease.
<span>During complete combustion, the hydrocarbon reacts with oxygen (O2) to from carbon dioxide (CO2) and water (H2O). For the combustion of one molecule of octane, 8 molecules of CO2 and 9 molecules of H2O must be formed to account for all the atoms of carbon and hydrogen in the octane. This requires 25 atoms of oxygen, or 12.5 molecules of O2:
C8H18+12.5*O2=9*H2O+8*CO2
Multiply both sides by 2 to obtain whole number coefficients:
2*C8H18+25*O2=18*H2O+16*CO2
The coefficient of carbon dioxide in the balanced equation is 16.</span>
Amobarbital (like all barbiturates) works by being incontestible to the GABAA receptor at either the alpha or the beta subunit.
<h3>What is the mechanism of amobarbital?</h3>
Amobarbital (like all barbiturates) works by binding to the GABAA receptor at either the alpha or the beta subunit. These are compulsory sites that are distinct from GABA itself and also distinct from the benzodiazepine binding site.
Amobarbital is a barbiturate classified as having a halfway duration of action, meaning that the effects of the drug can last from 4-6 amobarbital increases the effects of benazepril by apparatus: pharmacodynamic synergism.
So we can conclude that Amobarbital, 5-ethyl-5-isoamyl barbituric acid like all barbiturates.
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From the calculations, the pH of the final solution is 9.04.
<h3>What is the pH of the buffer?</h3>
We can use the Henderson Hasselbach equation to obtain the final pH of the solution in terms of the pKb and the base concentration.
Number of moles of salt = 250/1000 L * 0.5 M = 0.125 moles
Number of moles of base = 150/1000 L * 0.5 M = 0.075 moles
Total volume of solution = 250ml + 150ml = 400ml or 0.4 L
Molarity of base = 0.075 moles/ 0.4 L = 0.1875 M
Molarity of salt = 0.125 moles/ 0.4 L = 0.3125 M
pOH = pKb + log[salt/base]
pKb = -log(1.8 x 10^-5) = 4.74
pOH = 4.74 + log[0.3125/0.1875 ]
pOH = 4.96
pH = 14- 4.96
pH = 9.04
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